Pressure is defined as the force per unit area:

\[P = \frac{F}{A}.\]

If an object is immersed in a liquid at a depth \(h\), the fluid pressure is given by the constant depth formula

\[P = \rho gh,\]

where \(\rho\) is the fluid density and \(g\) is the acceleration due to gravity.

Fluid pressure is a scalar quantity. It has no direction, so a fluid exerts pressure equally in all directions. This statement is known as Pascal’s law discovered by the French scientist Blaise Pascal (\(1623 – 1662\)).

Consider the case when a vertical plate bounded by the lines

\[{x = a,\;\;}\kern0pt{x = b,\;\;}\kern0pt{y = f\left( x \right),\;\;}\kern0pt{y = g\left( x \right)}\]

is immersed in a liquid.

Since the different points of the lamina are at different depths, the total hydrostatic force \(F\) acting on the lamina is determined through integration:

\[F = \rho g\int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]xdx} .\]

This formula is often referred as the variable depth formula for fluid force.

## Solved Problems

Click a problem to see the solution.

### Example 1

A cylindrical tank with height of \(3\,\text{m}\) and base radius \(1\,\text{m}\) is filled with gasoline. Calculate the hydrostatic force exerted on the wall of the tank if the density of gasoline is \(800\,\large{\frac{{\text{kg}}}{{{\text{m}^3}}}}\normalsize.\)### Example 2

A rectangular swimming pool is \(H\) meters deep, \(a\) meters wide, and \(b\) meters long. Calculate- The fluid force \(F_{ab}\) acting on the bottom of the pool;
- The fluid force \(F_{aH}\) acting on each \(\left({a \times H}\right)\text{m}\) side;
- The fluid force \(F_{bH}\) acting on each \(\left({b \times H}\right)\text{m}\) side;

### Example 3

A triangular plate with base \(a\) and height \(H\) is submerged vertically in water so that its base lies at the surface of the water. Find the hydrostatic force acting on each side of the plate.### Example 4

A cube with side \(a\) is submerged in water so that its top face is parallel to the water surface and \(H\) meters below it. Find the total hydrostatic force acting on the cube.### Example 5

A rectangular plate with sides \(a\) and \(b\) \(\left({a \gt b}\right)\) is submerged in water at an angle \(\alpha\) to the water surface. The longer side is parallel to the surface and lies at a depth \(H\). Find the force acting on each side of the plate.### Example 6

A dam has the shape of an isosceles trapezoid with upper base \(a = 64\,\text{m},\) lower base \(b = 42\,\text{m},\) and height \(H = 3\,\text{m}.\) Find the force on the dam due to hydrostatic pressure.### Example 7

A right circular cone with base radius \(R\) and altitude \(H\) is submerged, vertex downwards, in water so that its base is on the surface of the water. Find the force due to hydrostatic pressure acting on the lateral cone surface.### Example 8

A plate in the shape of a parallelogram with sides \(a, b\) and angle \(\alpha\) is submerged vertically in water, so that the side \(b\) is at the water surface. Calculate the hydrostatic force acting on each side of the plate.### Example 9

A disk of radius \(R\) is half submerged vertically in liquid of density \(\rho.\) Find the hydrostatic force acting on one side of the disk.### Example 10

A plate in the shape of a parabolic segment is submerged vertically in water as shown in Figure \(12.\) The base of the segment is \(2a,\) the height is \(H.\) Find the force due to hydrostatic pressure acting on each side of the plate.### Example 1.

A cylindrical tank with height of \(3\,\text{m}\) and base radius \(1\,\text{m}\) is filled with gasoline. Calculate the hydrostatic force exerted on the wall of the tank if the density of gasoline is \(800\,\large{\frac{{\text{kg}}}{{{\text{m}^3}}}}\normalsize.\)Solution.

We choose the \(x-\)axis directed vertically downward with origin at the top base of the tank.

Consider a thin layer at a depth of \(x.\) If its thickness is \(dx,\) the lateral surface area of the layer is given by

\[dA = 2\pi Rdx.\]

The fluid pressure at the depth \(x\) is \(P = \rho gx,\) so the force exerted by the fluid on the lateral surface is

\[dF = PdA = 2\pi \rho gRxdx.\]

To find the total hydrostatic force \(F,\) we integrate from \(x = 0\) to \(x = H:\)

\[\require{cancel}{F = \int\limits_0^H {dF} }={ 2\pi \rho gR\int\limits_0^H {xdx} }={ \left. {\frac{{\cancel{2}\pi \rho gR{x^2}}}{\cancel{2}}} \right|_0^H }={ \left. {\pi \rho gR{x^2}} \right|_0^H }={ \pi \rho gR{H^2}.}\]

Substituting the given values into the formula, we have

\[{F = \pi \times 800 \times 9.8 \times 1 \times {3^2} }\approx{ 221671\,\text{N} }\approx{ 222\,\text{kN}.}\]

### Example 2.

A rectangular swimming pool is \(H\) meters deep, \(a\) meters wide, and \(b\) meters long. Calculate- The fluid force \(F_{ab}\) acting on the bottom of the pool;
- The fluid force \(F_{aH}\) acting on each \(\left({a \times H}\right)\text{m}\) side;
- The fluid force \(F_{bH}\) acting on each \(\left({b \times H}\right)\text{m}\) side;

Solution.

- The pressure at the bottom of the swimming pool is \(P = \rho gH,\) so the hydrostatic force acting on the bottom is given by
\[{F_{ab}} = PA = \rho gHA = \rho gabH.\]

- To determine the force on the \(\left({a \times H}\right)\text{m}\) side of the pool, we take a thin strip of thickness \(dx\) at a depth \(x.\)
The area of the strip is \(dA = adx.\) Since the water pressure at depth \(x\) is \(P = \rho gx,\) the force acting on the elementary strip is
\[dF = PdA = \rho gaxdx.\]

The total force on the \(\left({a \times H}\right)\text{m}\) side is obtained by integration:\[{F_{aH} = \int\limits_0^H {dF} }={ \int\limits_0^H {dF} }={ \rho ga\int\limits_0^H {xdx} }={ \left. {\frac{{\rho ga{x^2}}}{2}} \right|_0^H }={ \frac{{\rho ga{H^2}}}{2}.}\]

- Similarly we can find the force acting on \(\left({b \times H}\right)\text{m}\) side of the pool:
\[{F_{bH}} = \frac{{\rho gb{H^2}}}{2}.\]

### Example 3.

A triangular plate with base \(a\) and height \(H\) is submerged vertically in water so that its base lies at the surface of the water. Find the hydrostatic force acting on each side of the plate.Solution.

From similar triangles we have

\[{\frac{W}{a} = \frac{{H – x}}{H},}\;\; \Rightarrow {W = a – \frac{a}{H}x.}\]

The area of the elementary horizontal strip at depth \(x\) is

\[{dA = Wdx }={ \left( {a – \frac{a}{H}x} \right)dx.}\]

The water pressure at depth \(x\) is \(P = \rho gx,\) so the force acting on the strip is written as

\[{dF = PdA }={ \rho gx\left( {a – \frac{a}{H}x} \right)dx }={ \rho gax\left( {1 – \frac{x}{H}} \right)dx.}\]

The total force is determined through integration:

\[{F = \int\limits_0^H {dF} }={ \rho ga\int\limits_0^H {x\left( {1 – \frac{x}{H}} \right)dx} }={ \rho ga\int\limits_0^H {\left( {x – \frac{{{x^2}}}{H}} \right)dx} }={ \rho ga\left. {\left[ {\frac{{{x^2}}}{2} – \frac{{{x^3}}}{{3H}}} \right]} \right|_0^H }={ \rho ga\left( {\frac{{{H^2}}}{2} – \frac{{{H^3}}}{{3H}}} \right) }={ \frac{{\rho ga{H^2}}}{6}.}\]

### Example 4.

A cube with side \(a\) is submerged in water so that its top face is parallel to the water surface and \(H\) meters below it. Find the total hydrostatic force acting on the cube.Solution.

Using the the constant depth formula, it is easy to find the force acting on the top face:

\[{F_{top}} = {P_{top}}A = \rho g{a^2}H.\]

Similarly, the force on the bottom face is written as

\[{{F_{bottom}} = {P_{bottom}}A }={ \rho g{a^2}\left( {H + a} \right) }={ \rho g{a^2}H + \rho g{a^3}.}\]

To determine the side force, we consider a thin horizontal strip of thickness \(dx\) at depth \(x.\) Its area is \(dA = adx.\) The water pressure at this depth is \(P = \rho gx,\) so the hydrostatic force \(dF\) acting on the strip is given by the expression

\[dF = PdA = \rho gaxdx.\]

Then the force acting on the entire face of the cube is obtained by integration:

\[{{F_{side}} = \int\limits_H^{H + a} {dF} }={ \rho ga\int\limits_H^{H + a} {xdx} }={ \left. {\frac{{\rho ga{x^2}}}{2}} \right|_H^{H + a} }={ \frac{{\rho ga}}{2}\left[ {{{\left( {H + a} \right)}^2} – {H^2}} \right] }={ \frac{{\rho ga}}{2}\left( {\cancel{{H^2}} + 2aH + {a^2} -\cancel{{H^2}}} \right) }={ \rho g{a^2}H + \frac{{\rho g{a^3}}}{2}.}\]

The total hydrostatic force acting on the cube is given by

\[{F = {F_{top}} + {F_{bottom}} + 4{F_{side}} }={ \rho g{a^2}H + \rho g{a^2}H }+{ \rho g{a^3} }+{ 4\left( {\rho g{a^2}H + \frac{{\rho g{a^3}}}{2}} \right) }={ 6\rho g{a^2}H + 3\rho g{a^3} }={ 3\rho g{a^2}\left( {2H + a} \right).}\]

### Example 5.

A rectangular plate with sides \(a\) and \(b\) \(\left({a \gt b}\right)\) is submerged in water at an angle \(\alpha\) to the water surface. The longer side is parallel to the surface and lies at a depth \(H\). Find the force acting on each side of the plate.Solution.

By Pascal’s law, the fluid pressure at a depth \(x\) is \(P = \rho gx\) in any direction. So if we take a small strip on the plate at depth \(x\) corresponding to the increment \(dx,\) the force acting on the strip is given by

\[{dF = PdA }={ \rho gx \times \frac{{adx}}{{\sin \alpha }} }={ \frac{{\rho gaxdx}}{{\sin \alpha }}.}\]

To total hydrostatic force is obtained by integration:

\[{F = \int\limits_H^{H + b\sin \alpha } {dF} }={ \frac{{\rho ga}}{{\sin \alpha }}\int\limits_H^{H + b\sin \alpha } {xdx} }={ \frac{{\rho ga}}{{\sin \alpha }}\left. {\frac{{{x^2}}}{2}} \right|_H^{H + b\sin \alpha } }={ \frac{{\rho ga}}{{2\sin \alpha }}\left[ {{{\left( {H + b\sin \alpha } \right)}^2} – {H^2}} \right] }={ \frac{{\rho ga}}{{2\sin \alpha }}\left( {2bH\sin \alpha + {b^2}{{\sin }^2}\alpha } \right) }={ \rho gab\left( {H + \frac{b}{2}\sin \alpha } \right).}\]

### Example 6.

A dam has the shape of an isosceles trapezoid with upper base \(a = 64\,\text{m},\) lower base \(b = 42\,\text{m},\) and height \(H = 3\,\text{m}.\) Find the force on the dam due to hydrostatic pressure.Solution.

If we choose the vertical \(x−\)axis directed downward, the fluid pressure at a depth \(x\) is written as

\[P = \rho gx.\]

A thin horizontal strip of width \(dx\) at depth \(x\) can be approximated by a rectangle with the area equal to

\[dA = Wdx,\]

where the width \(W\) of the trapezoid at depth \(x\) is determined from similar triangles and is given by

\[W = a – \left( {a – b} \right)\frac{x}{H}.\]

Hence, the hydrostatic force acting on the strip is expressed by the formula

\[{dF = PdA }={ \rho gx\left[ {a – \left( {a – b} \right)\frac{x}{H}} \right]dx.}\]

The total force exerted on the dam due to hydrostatic pressure is given by

\[{F = \int\limits_0^H {dF} }={ \rho g\int\limits_0^H {x\left[ {a – \left( {a – b} \right)\frac{x}{H}} \right]dx} }={ \rho g\int\limits_0^H {\left( {ax – \frac{{a – b}}{H}{x^2}} \right)dx} }={ \rho g\left. {\left[ {\frac{{a{x^2}}}{2} – \frac{{\left( {a – b} \right){x^3}}}{{3H}}} \right]} \right|_0^H }={ \rho g\left[ {\frac{{a{H^2}}}{2} – \frac{{\left( {a – b} \right){H^2}}}{3}} \right] }={ \rho g{H^2}\left( {\frac{a}{6} + \frac{b}{3}} \right).}\]

Now we can easily calculate the value of the force:

\[{F = 1000 \times 9.8 \times {3^2} \times \left( {\frac{{6.4}}{6} + \frac{{4.2}}{3}} \right) }={ 217560\,\text{N} }\approx{ 218\,\text{kN}.}\]

### Example 7.

A right circular cone with base radius \(R\) and altitude \(H\) is submerged, vertex downwards, in water so that its base is on the surface of the water. Find the force due to hydrostatic pressure acting on the lateral cone surface.Solution.

We have the following proportion from similar triangles:

\[{\frac{W}{{H – x}} = \frac{R}{H},}\;\; \Rightarrow {W = \frac{{R\left( {H – x} \right)}}{H} = R\left( {1 – \frac{x}{H}} \right).}\]

The surface area of the elementary cone strip at the point \(x\) is given by

\[{dA = 2\pi Wdx }={ 2\pi R\left( {1 – \frac{x}{H}} \right)dx.}\]

The pressure in any direction at depth \(x\) is \(P = \rho gx,\) so the force on the strip is equal to

\[{dF = PdA }={ 2\pi \rho gRx\left( {1 – \frac{x}{H}} \right)dx.}\]

The total force is obtained by integrating from \(x = 0\) to \(x = H:\)

\[{F = \int\limits_0^H {dF} }={ 2\pi \rho gR\int\limits_0^H {x\left( {1 – \frac{x}{H}} \right)dx} }={ 2\pi \rho gR\int\limits_0^H {\left( {x – \frac{{{x^2}}}{H}} \right)dx} }={ 2\pi \rho gR\left. {\left( {\frac{{{x^2}}}{2} – \frac{{{x^3}}}{{3H}}} \right)} \right|_0^H }={ 2\pi \rho gR\left( {\frac{{{H^2}}}{2} – \frac{{{H^3}}}{{3H}}} \right) }={ \frac{{\pi \rho gR{H^2}}}{3}.}\]

### Example 8.

A plate in the shape of a parallelogram with sides \(a, b\) and angle \(\alpha\) is submerged vertically in water, so that the side \(b\) is at the water surface. Calculate the hydrostatic force acting on each side of the plate.Solution.

The vertices of the parallelogram \(ABCD\) are

\[{A\left( {0,0} \right),\;\;}\kern0pt{B\left( {0,b} \right),\;\;}\kern0pt{C\left( {a\sin \alpha ,b + a\cos \alpha } \right),\;\;}\kern0pt{D\left( {a\sin \alpha ,a\cos \alpha } \right).}\]

Determine the equation of the side \(AD.\) Using the two-point form of straight line equation, we have:

\[{\frac{{x – {x_A}}}{{{x_D} – {x_A}}} = \frac{{y – {y_A}}}{{{y_D} – {y_A}}},}\;\; \Rightarrow {\frac{{x – 0}}{{a\sin \alpha – 0}} = \frac{{y – 0}}{{a\cos \alpha – 0}},}\;\; \Rightarrow {\frac{x}{{a\sin \alpha }} = \frac{y}{{a\cos \alpha }},}\;\; \Rightarrow {{y_1} = x\cot \alpha .}\]

The other side \(BC\) is shifted \(b\) units upwards along the \(y-\)axis, so its equation is given by

\[{y_2} = b + x\cot \alpha .\]

Now we use we apply the variable depth formula:

\[F = \rho g\int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]xdx} .\]

This gives the total force acting on the plate:

\[{F = \rho g\int\limits_0^{a\sin \alpha } {\left( {{y_2} – {y_1}} \right)xdx} }={ \rho g\int\limits_0^{a\sin \alpha } {\left( {b + \cancel{x\cot \alpha} – \cancel{x\cot \alpha} } \right)xdx} }={ \rho gb\int\limits_0^{a\sin \alpha } {xdx} }={ \left. {\frac{{\rho gb{x^2}}}{2}} \right|_0^{a\sin \alpha } }={ \frac{{\rho gb{a^2}{{\sin }^2}\alpha }}{2}.}\]

### Example 9.

A disk of radius \(R\) is half submerged vertically in liquid of density \(\rho.\) Find the hydrostatic force acting on one side of the disk.Solution.

Consider a thin horizontal strip of thickness \(dx\) at a depth \(x.\) The width of the strip is

\[W = AB = 2\sqrt {{R^2} – {x^2}},\]

so its area is

\[{dA = Wdx }={ 2\sqrt {{R^2} – {x^2}} dx.}\]

The force on the strip is approximately

\[{dF = PdA }={ \rho gxdA }={ 2\rho gx\sqrt {{R^2} – {x^2}} dx.}\]

The total hydrostatic force is given by the integral

\[{F = \int\limits_0^R {dF} }={ 2\rho g\int\limits_0^R {x\sqrt {{R^2} – {x^2}} dx} .}\]

We evaluate this integral using the change of variable:

\[{I = \int {x\sqrt {{R^2} – {x^2}} dx} }={ \left[ {\begin{array}{*{20}{l}} {z = {R^2} – {x^2}}\\ {dz = – 2xdx} \end{array}} \right] }={ \int {\sqrt z \left( { – \frac{{dz}}{2}} \right)} }={ – \frac{1}{2}\int {\sqrt z dz} }={ – \frac{{{z^{\frac{3}{2}}}}}{3} }={ – \frac{{\sqrt {{z^3}} }}{3} }={ – \frac{{\sqrt {{{\left( {{R^2} – {x^2}} \right)}^3}} }}{3}.}\]

Hence, the force \(F\) is given by

\[{F = – \frac{{2\rho g}}{3}\left. {\sqrt {{{\left( {{R^2} – {x^2}} \right)}^3}} } \right|_0^R }={ – \frac{{2\rho g}}{3}\left( {0 – {R^3}} \right) }={ \frac{{2\rho g{R^3}}}{3}.}\]

### Example 10.

A plate in the shape of a parabolic segment is submerged vertically in water as shown in Figure \(12.\) The base of the segment is \(2a,\) the height is \(H.\) Find the force due to hydrostatic pressure acting on each side of the plate.Solution.

First we determine the equation of the parabola given its base \(2a\) and height \(H.\) The initial equation is \(x = H – k{y^2}.\) Since \(y = a\) at the point \(x = 0, \) the coefficient \(k\) is equal to

\[{0 = H – k{a^2},}\;\; \Rightarrow {k = \frac{H}{{{a^2}}}.}\]

This yields:

\[{x = H – \frac{H}{{{a^2}}}{y^2} }={ H\left( {1 – \frac{{{y^2}}}{{{a^2}}}} \right).}\]

By solving this equation for \(y,\) we get

\[{\frac{x}{H} = 1 – \frac{{{y^2}}}{{{a^2}}},}\;\; \Rightarrow {{a^2} – {y^2} = {a^2}\frac{x}{H},}\;\; \Rightarrow {{y^2} = {a^2}\left( {1 – \frac{x}{H}} \right).}\]

So, the parabola segment is bounded by the curves

\[{y = g\left( x \right) = – a\sqrt {1 – \frac{x}{H}} ,\;\;}\kern0pt{y = f\left( x \right) = a\sqrt {1 – \frac{x}{H}} .}\]

To calculate the hydrostatic force, we apply the variable depth formula:

\[F = \rho g\int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]xdx} .\]

In our case,

\[F = 2\rho ga\int\limits_0^H {\sqrt {1 – \frac{x}{H}} xdx} .\]

Make the substitution

\[{1 – \frac{x}{H} = {z^2},}\;\; \Rightarrow {x = H(1 – {z^2}),\;\;}\kern0pt{dx = – 2Hzdz.}\]

When \(x = 0,\) \(z = 1,\) and when \(x = H,\) \(z = 0.\) Hence

\[{F = – 4\rho ga{H^2}\int\limits_1^0 {z\left( {1 – {z^2}} \right)zdz} }={ 4\rho ga{H^2}\int\limits_0^1 {\left( {{z^2} – {z^4}} \right)dz} }={ 4\rho ga{H^2}\left. {\left( {\frac{{{z^3}}}{3} – \frac{{{z^5}}}{5}} \right)} \right|_0^1 }={ 4\rho ga{H^2}\left( {\frac{1}{3} – \frac{1}{5}} \right) }={ \frac{{8\rho ga{H^2}}}{{15}}.}\]